A) a parabola whose focus is\[\left( \frac{1}{4},0 \right)\]and directrix is\[x=-\frac{1}{4}\]
B) a parabola whose vertex is (3, 4) and directrix is\[x=\frac{11}{4}\]
C) a parabola whose focus is\[\left( \frac{13}{4},4 \right)\]and vertex is (0, 0)
D) a curve which is not a parabola
Correct Answer: B
Solution :
Given equation is \[{{y}^{2}}-8y-x+19=0\] \[\Rightarrow \] \[{{(y-4)}^{2}}=x-19+16\] \[\Rightarrow \] \[{{(y-4)}^{2}}=(x-3)\] \[\Rightarrow \] \[{{y}^{2}}=4AX\] where\[Y=y-4,A=\frac{1}{4}\]and\[X=x-3\] \[\therefore \]Focus\[=(A,0)=\left( \frac{1}{4},0 \right)=\left( \frac{13}{4},4 \right)\] Vertex = (3, 4) Directrix, \[X=-\frac{1}{4}\] \[\Rightarrow \] \[x-3=-\frac{1}{4}\] \[\Rightarrow \] \[x=\frac{11}{4}\]
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